# Can someone PROVE hero’s (or Heron’s) formula?

**Answer**

We use the formula area = (1/2)bc.sin(A)

The formula above can be “proven” easily by definition that a triangle’s area is just 1/2 x “base” x “height”.

Note in above, base = b. and height = c sinA.

Its one way of looking at it. Draw a diagram.

We’ll use that and some basic trigonometry to prove “Hero’s formula”.

And sin(A) = 2 sin(A/2).cos(A/2)

So area = bc.sin(A/2).cos(A/2)

Next we find expressions for cos(A/2) and sin(A/2) in terms of

s, a, b, c, where a, b, c are the sides of the triangle and s is

the semi-perimeter.

So s = (a+b+c)/2

cos(A)

= 2cos^2(A/2) – 1

= [ b^2 + c^2 – a^2 ] / [2bc]

So,

2cos^2(A/2)

= 1 + [ b^2 + c^2 – a^2 ] / [2bc]

= [(b+c)^2 – a^2 ]/[2bc] = [(b+c-a)(b+c+a)]/[2bc]

= [(2s-2a)2s ]/[2bc]

So,

cos^2(A/2) = [s(s-a)] / [bc]

b^2 + c^2 – a^2

Next,

cos(A)

= 1 – 2 sin^2(A/2)

= [b^2 + c^2 – a^2] / [2bc]

So,

2sin^2(A/2) = 1 – [b^2 + c^2 – a^2] / [2bc]

2sin^2(A/2) = [a^2 – (b-c)^2] / [2bc]

2sin^2(A/2) = [(a-b+c)(a+b-c)] / [2bc]

2sin^2(A/2) = [(2s-2b)(2s-2c)] / [2bc]

Hence,

sin^2(A/2) = [(s-b)(s-c)] / [bc]

We can now return to the formula for the area of the triangle using the above:

(1/2)bc.sin(A) = bc.sin(A/2).cos(A/2)

(1/2)bc.sin(A) = [bc.sqrt[s(s-a)(s-b(s-c)]/[bc]

(1/2)bc.sin(A) = sqrt[s(s-a)(s-b)(s-c)]

Done.

Note : The dots “.” mean nothing just “spacing”.